Question

log a²/bc + log b²/ca + log c²/ab=0​

Answer 1

Step-by-step explanation:

NOTE :

log (a*b) = loga + logb

log (a/b) = loga - logb

logm^2 = 2logm

SOLUTION :

log a²/bc + log b²/ca + log c²/ab=0

loga^2 - logbc + logb^2 - logca + logc^2- logab =0

=> 2loga-(logb + logc) +2logb-(logc+loga)+ 2logc-(loga+logb) =0

=> 2loga-logb-logc+2logb-logc-loga+2logc-loga-logb = 0

Rearange same alphabet

=> 2loga-loga-loga+2logb-logb-logb+2logc-logc-logc = 0

=> 2loga - 2loga + 2logb - 2logb + 2logc -2logc=0

everything get cancelled inleft hand side.

Hence it is proved that

log a²/bc + log b²/ca + log c²/ab=0